# 'Ambiguity' of dual vectors $\{dx^i\}$ in cotangent space in general relativity

The metric tensor is defined as:

$$g = g_{ij}dx^i \otimes dx^j,$$

where I used the summation convention. We often omit the tensor product sign $$\otimes$$ and just write this as:

$$g = g_{ij}dx^idx^j.$$

My question is about the line element which is also written as:

$$ds^2 = g_{ij}dx^i dx^j.$$

I will describe what I already know about the metric tensor and line element below.

Suppose we have a Riemannian manifold $$M$$, then we can define at every point $$p\in M$$ a metric. This give raise to the metric tensor field (so for every point $$p$$ we get a tensor:), so $$g_p: T_p M\times T_p M\rightarrow \mathcal{R}$$, where $$\mathcal{R}$$ are the real numbers. The coordinates $$\{x^i\}$$ are then local coordinates at $$p$$. From this notion we can define the length of a path $$\gamma : [a,b]\rightarrow M$$ with the property that $$\gamma(t)\in M$$ for $$t\in[a,b]$$. We can define the length of this path as:

$$L_\gamma = \int^b_a\sqrt{\pm g_{ij}\frac{dx^i}{dt}\frac{dx^j}{dt}}dt = \int^b_a\sqrt{\pm g_{ij}dx^idx^j}=\int^b_ads.$$

Where the $$\pm$$ ensures that we get a real solution.

From this we see that the line element can be defined as:

$$ds^2 = g_{ij}dx^idx^j.$$

The problem that I have with this, is that $$L_\gamma \in \mathcal{R}$$ is a real number. It now seems that we did not use the idea that $$dx^i$$ is a dual vector in the cotangent space defined at every point $$p \in M$$, but that we now regard those differentials as infinitesemal displacements (since only then they add up to a length).

I am not sure what happens here, I also read other posts but they seem to skip the point that the dual vectors $$dx^i$$ get an ambiguous meaning. For example if we consider the arc length in Cartesian coordinates and we let $$M=\mathcal{R}^2$$, then we get the following expression for the length of some curve:

$$L = \int^b_a \sqrt{dx^2+dy^2},$$

but how is this elementary notion of calculus consistent with the idea that $$dx^i$$ is a dual vector (so that $$dx^2$$ would actually be a tensor of rank (0,2))? Now it seems that everyone just assumes that the square root of the metric tensor is well-defined and that this evaluates to a scalar. Could somebody elaborate a bit more on this idea? The idea what a length and the metric tensor and so on is, is quite clear for me.

• @mavzolej please do not link to illegal copies of copyrighted materials. – Kyle Kanos Sep 30 '18 at 22:14
• @KyleKanos: We have no way of knowing what the copyright laws are where mavzolej lives. A more reasonable expectation might be that mavzolej note in the comment that the link is to a pirated copy of the book. The textbook market is so economically exploitative that I often make a point of mentioning to my students that Library Genesis exists. – user4552 Sep 30 '18 at 23:19
• @BenCrowell my understanding is that it doesn't matter where mavzolej lives, only that the servers on which we are all accessing are based in the US & so such copyright infringement laws in the US should be applied. I believe there are plenty of Mother Meta posts about it, if you're interested in learning about the position of SE on such thievery. – Kyle Kanos Sep 30 '18 at 23:21

Given a curve $$\gamma:I\to M$$, one can pull back the (positive definite) metric tensor $${\bf g}~=~ g_{ij}\mathrm{d}x^i \odot \mathrm{d}x^i~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}M)\right)\tag{A}$$ to $$\omega\odot\omega~:=~\gamma^{\ast}{\bf g}~\in~ \Gamma\left( {\rm Sym}^2(T^{\ast}I)\right),\tag{B}$$ where the 1-form $$\omega$$ is locally exact $$\omega~=~\mathrm{d}s\tag{C}$$ because of Poincare lemma $$\mathrm{d}\omega=0$$. Now to get a function on both sides of eq. (B) insert a vector field $$X\in \Gamma(TI)$$ twice on both sides of eq. (B). This will essentially reproduce what physicists mean by the formula $$ds^2 ~=~g_{ij}dx^idx^j. \tag{D}$$
The mismatch with the notations $$g= g_{ij}dx^i\otimes dx^j$$ and $$ds^2 = g_{ij}dx^idx^j$$ is due to the fact that, in this second formula the $$dx^k$$s correspond to "small" components of contravariant vectors $$\delta X$$ (the $$dx^k$$s are real numbers here) instead of elements of the dual basis (the $$dx^i$$ are covariant vectors in the rigorous view). Replacing in the second formula $$dx^k$$ with a less ambiguous notation $$\delta x^k$$, $$\delta X = \delta x^k \frac{\partial}{\partial x^k}$$ Therefore we have the identity connecting the two notions $$g_{ij} \delta x^i \delta x^j = g(\delta X, \delta X) = \delta s^2\:.$$ This interpretation is also in agreement with the integral defining the length of a curve provided we write $$\dot{\gamma} dt = \delta x^i \frac{\partial}{\partial x^i}$$, so that the formula for computing the length of a curve becomes $$L_\gamma = \int_a^b \sqrt{|g(\dot{\gamma},\dot{\gamma})|} dt = \int \sqrt{|g_{ij}\delta x^i\delta x^j|}\:.$$ The point is that the objects $$\delta x^k$$ do not exist as "infinitesimal numbers" (unless taking advantage of the nonstandard analysis) while the objects $$dx^i$$ do exist as vectors in a suitable vector space and this notation is therefore more rigorous from the mathematical viewpoint.